# Boundary conditions for eigenfunctions in a potential step

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1. Sep 3, 2017

### Dario SLC

1. The problem statement, all variables and given/known data
A particle with mass m and spin 1/2, it is subject in a spherical potencial step with height $V_0$.
How is the general form for the eigenfunctions?
What is the boundary conditions for this eigenfunctions?
Find the degeneracy level for the energy, when it is $E<V_0$

2. Relevant equations

-\frac{\hbar^2}{2m}\left(\frac{\partial^2u}{\partial r^2}-\frac{l(l+1)}{r^2}u\right)+V(r)u(r)=Eu(r)

with $u(r)=rR(r)$

3. The attempt at a solution
Well, first the constant of motion it $\hat{L_z}$, $\hat{L^2}$ and $\hat{S_z}$ because conmute with $\hat{H}$, also $\hat{H}$, and it former a complete set of observables that conmute, therefore the form of eigenfunctions are:

\psi(r,\theta,\varphi,\sigma)=R(r)Y(\theta,\varphi)\chi(\sigma)

when $\chi(\sigma)$ is a spin function, and only take the values $m_s={+1/2,-1/2}$ because eigenvalues of operator $\hat{S^2}$ is $3/4\hbar^2$ with $s=1/2$.

Here my doubt:
Boundary conditions to $Y(\theta,\varphi)$ this is referred to:
$l=0, 1, \dots, n-1$ and $-l\leq m\leq +l$?

For radial function, it must be:
because $\hat{H}$ is hermitian, then for $r=0\Longrightarrow u(0)\longrightarrow0$
and for $r\longrightarrow\infty \Longrightarrow u\longrightarrow0$

For $\chi$ the boundary conditions it is $-s\leq m_s\leq+s$?

In the case for the degeneracy, when $E<0$ corresponds to $l=0$ because it is a bound state, then without spin the level of degeneracy is $n^2$ but with spin is $2n^2$.

Up to here my poor attemp to solution, i am really puzzled for explain the boundary condition to Y and $\chi$, and i am not sure that the
$r\longrightarrow\infty \Longrightarrow u\longrightarrow0$ for this potential.

Thanks a lot!

2. Sep 8, 2017

### PF_Help_Bot

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